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1 1. Let A, B and C are three events such that PA =.4, PB =.3, PC =.3, P A B =.6, P A C =.6, P B C =., P A B C =.7. a Compute P A B, P A C, P B C. b Compute P A B C. c Compute the probability that exactly one of A, B and C happens. Solution. a P A B = P A + P B P A B =.1, P A C = P A + P C P A C =., P B C = P B + P C P B C =.1. b Since P A B C = P A + P B + P C P A B P A C P B C + P A B C we have P A B C = P A B C P A P B P C+P A B+P A C+P B C =.1. c We have P A B C = P A B P A B C =., P A B C = P A C P A B C =.1, P A B C = P B C P A B C =.. Now P A B C = P A P A B C P A B C P A B C =.. P A B C = P B P A B C P A B C P A B C =.1. P A B C = P C P A B C P A B C P A B C =.. So the answer is =.4.. You are dealt a -card hand from a well-shuffled deck of. a What is the probability of the union of the event that both cards are aces with the event that both cards are redhearts or diamons? b That is the probability that both cards have the same denomination e.g. both are s or both are jacks? c What is the conditional probability that the second card is a picture card J, Q, or K given that at least one of the two cards is a picture card? Solution. a Let A be the event that both cards are aces, and R be the event that both 4 6 cards are red. Then P A = since there are 4 aces. Likewise P R = 1. Next

2 P AR = 1 since there is only one good combination A, A. Hence the answer is b After the first card is drawn there are 1 cards left in the desk and 3 cards have the same denomination as the first card. Hence the answer is 3. 1 c Let B 1, B be the events that the first respectively, second card is a picture card. There are 1= 3 4 picture cards. Hence 1 P B 1 = P B = 1.3 and P B 1B =.. Thus if B is an event that at least one card is a picture card, then Hence P B 1 B = P B 1 B P B = P B 1 P B.6. P B = P B 1 + P B P B 1 B In a certain city there are two food stores. 7% of the the population use Cheap FoodStore and 3% use Green FoodStore. Among the shoopers of Cheap FoodStore 7% have household income of less than per year while among the shoopers of Green FoodStore % have household income of less than per year. a Find the probability that a randomly chosen citizen uses Green FoodStore and has household income or more per year. b What proportion of the citizens have household income or more per year. c Given that a person has a household income or more per year how likely he is to shop at Green FoodStore. Solution. Let C be event that a citizen uses Cheap FoodStore and G be event that a citizen uses Green FoodStore. Let R be the event that a citizen has a household income of or more per year. Note that P R C = 1 P R C =.3 and P R G = 1 P R G =.. a P RG = P GP R G =.3. =.1. b P R = P RG + P RC = =.36. c P G R = P RG/P R = /1. 4. Let X have cumulative distribution function F X x = x for x 1 and F X x = for x < and = 1 for x > 1,

8 8 c Estimate true mean heart rate during laughter in a way that conveys information about precision and reliability. Calculate 9% CI. Solution. a We have to test H = {µ = 71} vs H a = {µ > 71}. Test statistics is z = X 71 s/. 9 In our case z =.1 3/ = 6.6. Since z > z 9. = 1.6 we have sufficient evidence to reject H. That is we have a strong evidence that laughter causes the the average heart rate to exceed 71 beats per minute. b using the formula on page 314 of the book we get 71 7 β = Φ z. + 3/ = Φ Hence the power is 1 β =.93. c The confidence interval has the form s X ± z. = 73.1 ± = 73.1 ±.6 = [7.48, 73.7]. n 9 The upper confidence bound has the form s X ± z. = 73.1 ± = 73.1 ±.6 = [7.48, 73.7]. n 9 1. A business journal investigation of the performance and timing of corporate acquisitions discovered that in a random sample of, 863 firms, 848 announced one or more acquisitions during the year. Let p be the proportion of firms which made one or more acquisitions during the year. a Give a point estimate for p; b Construct 9% confidence interval for p; c Construct 9% upper confidence bound for p. Solution. a ˆp = X n = =.96. b The confidence interval takes form ˆp + z./ 863 ˆp1 ˆp/863 + z ±z. / z./ z./863 b The upper confidence bound takes form ˆp + z.1/ z.1/863 + z.1 ˆp1 ˆp/863 + z.1 / z.1/863 =.96±.14 = [.8,.31]. =.96 ±.11 = Strategic placement of lobster traps is one of keys for a successful lobster fisherman. A study was conducted of the average distance separating traps by lobster fishermen. The trap-spacing measurements in meters for a sample of seven teams from Blue Sea fishing

9 cooperative and came with the following data: x = 81. and s = 1. Suppose that the trap-spacing has normal distribution. a Test the hypotheis that the average distance between the traps is at least 9m at the significance level α =.. b What is the problem with using the normal z statistic to find a confidence interval for the average distance between the traps? c Construct 9% confidence interval for the average distance between the traps. d One team from Blue Sea cooperative was not working during the day of test. Given 9% prediction interval for the distance between the traps used by the missing team. Solution. a We have to test H = {µ = 9} vs H a = {µ < 9}. The test statistics is t = s x 9 which in our case equals to.13. The ctritiacal value t.,6 =.447. Since /7 t < t.,6 we do not reject H. That is we do not have sufficient evidence that the average distance between the traps is in fact less that 9 m. b We do not know the standard deviation and since the number of observations is small we can not estimate it accurately enough to use z statistic. c The confidence interval takes form 81 ±.447 s /7 = [7.69, 91.31]. Note that 9 is inside the confidence interval which is in agreement with the fact that we did not reject H. d The prediction interval takes form 81 ±.447 s = [1.84, 11.16]. 7 9

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